Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $k = \dfrac{x^2 - 3x - 18}{x^2 - 6x} \div \dfrac{x + 3}{8x + 64} $
Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{x^2 - 3x - 18}{x^2 - 6x} \times \dfrac{8x + 64}{x + 3} $ First factor the quadratic. $k = \dfrac{(x + 3)(x - 6)}{x^2 - 6x} \times \dfrac{8x + 64}{x + 3} $ Then factor out any other terms. $k = \dfrac{(x + 3)(x - 6)}{x(x - 6)} \times \dfrac{8(x + 8)}{x + 3} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (x + 3)(x - 6) \times 8(x + 8) } { x(x - 6) \times (x + 3) } $ $k = \dfrac{ 8(x + 3)(x - 6)(x + 8)}{ x(x - 6)(x + 3)} $ Notice that $(x - 6)$ and $(x + 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 8\cancel{(x + 3)}(x - 6)(x + 8)}{ x(x - 6)\cancel{(x + 3)}} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $k = \dfrac{ 8\cancel{(x + 3)}\cancel{(x - 6)}(x + 8)}{ x\cancel{(x - 6)}\cancel{(x + 3)}} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $k = \dfrac{8(x + 8)}{x} ; \space x \neq -3 ; \space x \neq 6 $